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17.4% K_(2)SO_(4) solition at 27^(@)C is isotonic with 4% NaOHsolution at the same temperature. If NaOH is 100% ionised, What is the degree of innisation of K_(2)SO_(4) in aqueous solution ? |
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Answer» Solution :Calculation of molar concerntration of both the solution. `"Molar concentration of NaOH solution"("Mass of NaOH / Molar mass")/("Volume of solution solution in LITRES")` `=((4g)//(40G mol^(-1)))/((0.1 L))=1.0 mol^(-1)=1.0 M` molar concentration of `K_(2)SO_(4)` solution = `(Mass of K_(2)SO_(4)//Molar mass)/("Colume of solution in liters")` `=((17.4g)//(174h mol^(-1)))/((0.1L))=10 molL^(-1)=1.0 M` Calculation of Van't Hoff factor (i) for `K_(2)SO_(4)` Since the two SOLUTIONS are isotonic, they have same osmotic pressure. `alpha=1, n=2` `alpha=(i-1)/(n-1)or1=(i-1)/(2-1)ori=2,c=1 M` `pi_(NaOH)=iCRT=2xx(1M)xxRxxT` C=1 M, i=? ( to be calcutaed) `pi_(K_(2)SO_(4))=oCRT=i(1)xxRxxT` As the twpsp,itopm are oisptpmec. `pi_(NaOH) or 2xxRxxT=ixxRxxT or o=2` Calculation of dgree of ionisation of `K_(2)SO_(4)` `K_(2)SO_(4): alpha=(i-1)/(n-1)=(2-1)/(3-1)=1/2=0.5` The degree of ionisation of `K_(2)SO_(4)=0.5` or its percentage ionisation=o.5xx100=50% |
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